Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - Chapter Review Exercises - Page 163: 32


$$ y'= \frac{-19}{(4t-9)^2}.$$

Work Step by Step

Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$ Since $ y=\frac{3t-2}{4t-9}$, then using the quotient rule, the derivative $ y'$ is given by $$ y'= \frac{(4t-9)(3)-(3t-2)(4)}{(4t-9)^2}= \frac{-19}{(4t-9)^2}.$$
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